I have been given the number: ${206746}^{20 }$and the problem wants me to compute the last digit using modular arithmetic. How would I go about this?
I know that since the ones digit is 6, no matter how big the exponential is the last digit is 6, but I do not know how to prove it.
In short, work mod $10$. This clearly extracts the last digit. You can then show that $6^n \equiv 6 \pmod {10}$ for all $n \geq 1$ very easily, noting that $6\cdot 6 \equiv 6 \pmod{10}$.
If you wish to avoid explicitly using modular arithmetic, you can use that $$(10x+6)^n = 6^n + 10x \times \binom{n}{1} + \dots$$ whre every term but the first has a factor of $10$ contributed from $(10x)^k$. So it is enough to look only at the last digit of $6^n$, which is $6$ by induction.
You used to have the tag proof-writing on the question. If it was appropriate here is an inductive proof.
Consider any number $x=10n+6$ where $n\in\mathbb{N}$.
Then $x\equiv6\pmod{10}$
Asume $x^k\equiv6\pmod{10}$. Hence it can be written in the form $x^n=10m+6$ where $m\in\mathbb{N}$.
So $x^{k+1}=x^k\times x=(10m+6)\times(10n+6)=100mn+10m+10n+36\equiv6\pmod{10}$
So if it is true for $k$ then it is also true for $k+1$ and it is true for $k=1$ so by the principle of mathematical induction it is true for any $k\in\mathbb{N}$.
